The Stacks project

Lemma 54.16.8. Let $S = \mathop{\mathrm{Spec}}(R)$ be an affine Noetherian scheme. Let $X \to S$ be a proper morphism. Let $\mathcal{L}$ be an ample invertible sheaf on $X$. Let $E \subset X$ be an exceptional curve of the first kind. Then

  1. there exists a contraction $b : X \to X'$ of $E$,

  2. $X'$ is proper over $S$, and

  3. the invertible $\mathcal{O}_{X'}$-module $\mathcal{L}'$ is ample with $\mathcal{L}'$ as in Remark 54.16.6.

Proof. Let $n$ be the degree of $\mathcal{L}|_ E$ as in Lemma 54.16.7. Observe that $n > 0$ as $\mathcal{L}$ is ample on $E$ (Varieties, Lemma 33.44.14 and Properties, Lemma 28.26.3). After replacing $\mathcal{L}$ by a power we may assume $H^ i(X, \mathcal{L}^{\otimes e}) = 0$ for all $i > 0$ and $e > 0$, see Cohomology of Schemes, Lemma 30.17.1. Finally, after replacing $\mathcal{L}$ by another power we may assume there exist global sections $t_0, \ldots , t_ n$ of $\mathcal{L}$ which define a closed immersion $\psi : X \to \mathbf{P}^ n_ S$, see Morphisms, Lemma 29.39.4.

Set $\mathcal{M} = \mathcal{L}(nE)$. Then $\mathcal{M}|_ E \cong \mathcal{O}_ E$. Since we have the short exact sequence

\[ 0 \to \mathcal{M}(-E) \to \mathcal{M} \to \mathcal{O}_ E \to 0 \]

and since $H^1(X, \mathcal{M}(-E))$ is zero (by Lemma 54.16.7 and the fact that $n > 0$) we can pick a section $s_{n + 1}$ of $\mathcal{M}$ which generates $\mathcal{M}|_ E$. Finally, denote $s_0, \ldots , s_ n$ the sections of $\mathcal{M}$ we get from the sections $t_0, \ldots , t_ n$ of $\mathcal{L}$ chosen above via $\mathcal{L} \subset \mathcal{L}(nE) = \mathcal{M}$. Combined the sections $s_0, \ldots , s_ n, s_{n + 1}$ generate $\mathcal{M}$ in every point of $X$ and therefore define a morphism

\[ \varphi : X \longrightarrow \mathbf{P}^{n + 1}_ S \]

over $S$, see Constructions, Lemma 27.13.1.

Below we will check the conditions of Lemma 54.16.4. Once this is done we see that the Stein factorization $X \to X' \to \mathbf{P}^{n + 1}_ S$ of $\varphi $ is the desired contraction which proves (1). Moreover, the morphism $X' \to \mathbf{P}^{n + 1}_ S$ is finite hence $X'$ is proper over $S$ (Morphisms, Lemmas 29.44.11 and 29.41.4). This proves (2). Observe that $X'$ has an ample invertible sheaf. Namely the pullback $\mathcal{M}'$ of $\mathcal{O}_{\mathbf{P}^{n + 1}_ S}(1)$ is ample by Morphisms, Lemma 29.37.7. Observe that $\mathcal{M}'$ pulls back to $\mathcal{M}$ on $X$ (by Constructions, Lemma 27.13.1). Finally, $\mathcal{M} = \mathcal{L}(nE)$. Since in the arguments above we have replaced the original $\mathcal{L}$ by a positive power we conclude that the invertible $\mathcal{O}_{X'}$-module $\mathcal{L}'$ mentioned in (3) of the lemma is ample on $X'$ by Properties, Lemma 28.26.2.

Easy observations: $\mathbf{P}^{n + 1}_ S$ is Noetherian and $\varphi $ is proper. Details omitted.

Next, we observe that any point of $U = X \setminus E$ is mapped to the open subscheme $W$ of $\mathbf{P}^{n + 1}_ S$ where one of the first $n + 1$ homogeneous coordinates is nonzero. On the other hand, any point of $E$ is mapped to a point where the first $n + 1$ homogeneous coordinates are all zero, in particular into the complement of $W$. Moreover, it is clear that there is a factorization

\[ U = \varphi ^{-1}(W) \xrightarrow {\varphi |_ U} W \xrightarrow {\text{pr}} \mathbf{P}^ n_ S \]

of $\psi |_ U$ where $\text{pr}$ is the projection using the first $n + 1$ coordinates and $\psi : X \to \mathbf{P}^ n_ S$ is the embedding chosen above. It follows that $\varphi |_ U : U \to W$ is quasi-finite.

Finally, we consider the map $\varphi |_ E : E \to \mathbf{P}^{n + 1}_ S$. Observe that for any point $x \in E$ the image $\varphi (x)$ has its first $n + 1$ coordinates equal to zero, i.e., the morphism $\varphi |_ E$ factors through the closed subscheme $\mathbf{P}^0_ S \cong S$. The morphism $E \to S = \mathop{\mathrm{Spec}}(R)$ factors as $E \to \mathop{\mathrm{Spec}}(H^0(E, \mathcal{O}_ E)) \to \mathop{\mathrm{Spec}}(R)$ by Schemes, Lemma 26.6.4. Since by assumption $H^0(E, \mathcal{O}_ E)$ is a field we conclude that $E$ maps to a point in $S \subset \mathbf{P}^{n + 1}_ S$ which finishes the proof. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0C2M. Beware of the difference between the letter 'O' and the digit '0'.